(a) Find direction vectors \(\overrightarrow{AB} = 2\mathbf{j} + 4\mathbf{k}\) and \(\overrightarrow{AC} = \mathbf{i} - \mathbf{j} + \mathbf{k}\).

Calculate the normal to the plane ABC using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 1 & -1 & 1 \end{vmatrix} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k}\)

Substitute point \(A(1, 0, -2)\) into \(3x + 2y - z = d\) to find \(d = 5\).

(b) Find normal to plane ABD:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 0 & 0 & t+2 \end{vmatrix} = (2t+4)\mathbf{i}\)

Use dot product to find angle:

\(\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2t+4 \\ 0 \\ 0 \end{pmatrix} = \sqrt{14} \cos \theta\)

\(\theta = 36.7^\circ\) when \(t \neq -2\).

(c) Find common perpendicular:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ -1 & 1 & t+1 \end{vmatrix} = \begin{pmatrix} -4 \\ 2 \\ 1 \end{pmatrix}\)

Use formula for perpendicular distance:

\(\frac{l + 2}{\sqrt{t^2 - 2t + 6}} = \sqrt{2}\)

Solve \(t^2 - 8t + 8 = 0\) to find \(t = 6.83, \; t = 1.17\).